Problem 25 - University Physics with Modern Physics

Question

A sinusoidal wave traveling in the positive \(x\) -direction has a wavelength of \(12 \mathrm{~cm},\) a frequency of \(10.0 \mathrm{~Hz},\) and an amplitude of \(10.0 \mathrm{~cm}\). The part of the wave that is at the origin at \(t=0\) has a vertical displacement of \(5.00 \mathrm{~cm} .\) For this wave, determine the a) wave number, d) speed, b) period, e) phase angle, and c) angular frequency, f) equation of motion.

Step-by-Step Solution

Answer

Based on the given information and step-by-step solutions, we found the following properties of the sinusoidal wave: 1. Wave number (k) = 52.36 m⁻¹ 2. Period (T) = 0.1 s 3. Angular frequency (ω) = 20π rad/s 4. Speed (v) = 1.2 m/s 5. Phase angle (φ) = 30° Finally, the equation of motion for the given wave is: y(x, t) = 10.0 cm sin(52.36 m⁻¹x - (20π rad/s)t + 30°)

Step 1: Wave number

The wave number (denoted by \(k\)) is the spatial frequency of the wave. To find the wave number, we can use the following equation: $$k = \frac{2\pi}{\lambda}$$ Here, \(\lambda\) is the wavelength which is given as \(12 cm\). So, let's calculate the wave number: $$k = \frac{2\pi}{12\ \text{cm}} = \frac{2\pi}{0.12\ \text{m}} \approx 52.36\ \text{m}^{-1}$$

Step 2: Period

The period (denoted by \(T\)) is the time taken for one complete oscillation. It is related to the frequency (denoted by \(f\)) as follows: $$T = \frac{1}{f}$$ Here, the frequency is given as \(10.0 Hz\). So, let's calculate the period: $$T = \frac{1}{10.0\ \text{Hz}} = 0.1\ \text{s}$$

Step 3: Angular frequency

The angular frequency (denoted by \(\omega\)) is related to the frequency by the following equation: $$\omega = 2\pi f$$ We have the frequency (\(10.0 Hz\)). Let's calculate the angular frequency: $$\omega = 2\pi (10.0\ \text{Hz}) = 20\pi\ \text{rad/s}$$

Step 4: Speed

The wave speed (denoted by \(v\)) is the speed at which the wave travels in the medium. It can be found using the following equation: $$v = \lambda f$$ We have the wavelength (\(12 cm\)) and the frequency (\(10.0 Hz\)). Let's calculate the wave speed: $$v = (12\ \text{cm})(10.0\ \text{Hz}) = 1.2\ \text{m/s}$$

Step 5: Phase angle

To find the phase angle (denoted by \(\phi\)), we need to consider the vertical displacement at the origin when \(t=0\). Let's denote the vertical displacement as \(y\). The wave equation is given by: $$y(x, t) = A\sin(kx - \omega t + \phi)$$ At \(t=0\) and \(x=0\), \(y(0,0) = 5.00\ \text{cm}\). Let's substitute the values and solve for \(\phi\): $$5.00\ \text{cm} = (10.0\ \text{cm})\sin(\phi)$$ $$\phi = \arcsin\left(\frac{1}{2}\right)$$ $$\phi = 30^{\circ}$$

Step 6: Equation of motion

Now we have all the necessary parameters to write the equation of motion for the given wave. The wave equation is given by: $$y(x, t) = A\sin(kx - \omega t + \phi)$$ Substitute the values we found for \(A\), \(k\), \(\omega\), and \(\phi\): $$y(x, t) = 10.0\ \text{cm}\sin\left(52.36\ \text{m}^{-1}x - (20\pi\ \text{rad/s})t + 30^{\circ}\right)$$

Key Concepts

Wave Number

The wave number, denoted by \(k\), captures how many wave cycles fit into a unit length. It's a measure of spatial frequency. The higher the wave number, the more oscillations occur over a specific distance. You can calculate it using the formula:

• \(k = \frac{2\pi}{\lambda}\)

where \(\lambda\) represents the wavelength.
In our example, the wavelength is given as \(12\ \text{cm}\), or \(0.12\ \text{m}\). By plugging this value into the equation, we find the wave number to be approximately \(52.36\ \text{m}^{-1}\).
This essentially tells us that in every meter, the wave completes slightly over 52 cycles. Understanding the wave number helps in visualizing how "tight" the wave oscillations are within a given space.

Angular Frequency

Angular frequency, represented by the Greek letter \(\omega\), relates to how fast the wave oscillates in terms of radians per second. It provides information on how quickly each cycle of the wave is completed. To calculate it, we use:

• \(\omega = 2\pi f\)

where \(f\) stands for frequency.
Given a frequency of \(10.0\ \text{Hz}\) in our problem, the angular frequency outweighs the regular frequency because we use radians rather than simple cycles. With this frequency, you find that \(\omega = 20\pi\ \text{rad/s}\).
This confirms that in \(1\ \text{s}\), the wave traverses \(20\pi\) radians, indicating its rather brisk oscillation.

Wave Equation

The wave equation offers a mathematical depiction of the wave's oscillation. It describes how the wave's characteristics change over time and position. The general form of a sinusoidal wave equation is:

• \(y(x, t) = A\sin(kx - \omega t + \phi)\)

where:

• \(y(x, t)\) is the vertical displacement,
• \(A\) is the amplitude,
• \(k\) is the wave number,
• \(\omega\) is the angular frequency, and
• \(\phi\) is the phase angle.

In this exercise, substituting the known values:

• \(A = 10.0\ \text{cm}\),
• \(k = 52.36\ \text{m}^{-1}\),
• \(\omega = 20\pi\ \text{rad/s}\), and
• \(\phi = 30^{\circ}\)

results in the specific equation:

• \(y(x, t) = 10.0\ \text{cm}\sin(52.36\ \text{m}^{-1}x - 20\pi\ \text{rad/s} \cdot t + 30^{\circ})\)

This equation models the wave's behavior, showing how the wave's vertical displacement at any point \(x\) and time \(t\) can be calculated.

Phase Angle

The phase angle, denoted \(\phi\), indicates the initial angle of the wave at time \(t = 0\). It helps determine the wave's starting position in its cycle, crucial for matching experimental data in practical applications. The phase angle can shift the wave forwards or backwards along the position axis.
In our exercise, the initial displacement \(y(0, 0)\) is \(5.00\ \text{cm}\). We solve for \(\phi\) using:

• \(5.00\ \text{cm} = 10.0\ \text{cm} \cdot \sin(\phi)\)

resulting in \(\phi = 30^{\circ}\).
This means at \(t=0\), our wave starts from a position that is a third of the way through its cycle from the baseline. With \(\phi\), we can precisely shift the wave to fit measurable starting positions.